A function $$ f $$ and a point $$ P $$ are given. Let $$ heta $$ correspond to the direction of the directional derivative. Complete parts a. through $$ e $$. $$ f(x, y)=\sqrt{3+2 x^{2}+2 y^{2}}, P(\sqrt{3},-1) $$ a. Find the gradient and evaluate it at $$ P $$. The gradient at $$ P $$ is $$ \left\langle\frac{2 \sqrt{3}}{\sqrt{11}},-\frac{2}{\sqrt{11}} ight) $$. (Type exact answers, using radicals as needed.) b. Find the angles $$ heta $$ (with respect to the positive x-axis) associated with the directions of maximum increase, maximum decrease, and zero change. What angles are associated with the direction of maximum increase? $$ \square $$ (Type any angles in radians between 0 and $$ 2 \pi $$. Type an exact answer, using $$ \pi $$ as needed. Use a comma to separate answers as needed.)

Question

A function $$ f $$ and a point $$ P $$ are given. Let $$ 	heta $$ correspond to the direction of the directional derivative. Complete parts a. through $$ e $$. $$ f(x, y)=\sqrt{3+2 x^{2}+2 y^{2}}, P(\sqrt{3},-1) $$ a. Find the gradient and evaluate it at $$ P $$. The gradient at $$ P $$ is $$ \left\langle\frac{2 \sqrt{3}}{\sqrt{11}},-\frac{2}{\sqrt{11}}ight) $$. (Type exact answers, using radicals as needed.) b. Find the angles $$ 	heta $$ (with respect to the positive x-axis) associated with the directions of maximum increase, maximum decrease, and zero change. What angles are associated with the direction of maximum increase? $$ \square $$ (Type any angles in radians between 0 and $$ 2 \pi $$. Type an exact answer, using $$ \pi $$ as needed. Use a comma to separate answers as needed.)

UpStudy AI · Accepted Answer

1. We are given  
   $$
   f(x,y)=\sqrt{3+2x^2+2y^2}
   $$
   and the point  
   $$
   P\bigl(\sqrt{3},\,-1\bigr).
   $$

2. The gradient is given by  
   $$
   \nabla f(x,y)=\left\langle f_x(x,y),\,f_y(x,y)\right\rangle,
   $$
   where  
   $$
   f_x(x,y)=\frac{2x}{\sqrt{3+2x^2+2y^2}} \quad \text{and} \quad f_y(x,y)=\frac{2y}{\sqrt{3+2x^2+2y^2}}.
   $$

3. Evaluating at $$ P $$ we have
   $$
   3+2x^2+2y^2\big|_{x=\sqrt{3},\,y=-1} = 3+2(3)+2(1)= 3+6+2=11.
   $$
   Thus,
   $$
   f_x\bigl(\sqrt{3},-1\bigr)=\frac{2\sqrt{3}}{\sqrt{11}}, \quad f_y\bigl(\sqrt{3},-1\bigr)=\frac{2(-1)}{\sqrt{11}}=-\frac{2}{\sqrt{11}}.
   $$
   Therefore,  
   $$
   \nabla f\bigl(\sqrt{3},-1\bigr)=\left\langle \frac{2\sqrt{3}}{\sqrt{11}},\,-\frac{2}{\sqrt{11}}\right\rangle.
   $$

4. The **direction of maximum increase** is in the direction of the gradient. To find its angle $$\theta$$ (with respect to the positive $$x$$-axis), we note that the gradient vector components are:
   $$
   \left( \frac{2\sqrt{3}}{\sqrt{11}},\, -\frac{2}{\sqrt{11}} \right).
   $$
   The angle $$\theta$$ is given by
   $$
   \theta=\arctan\left(\frac{y\text{-component}}{x\text{-component}}\right)=\arctan\left(\frac{-\frac{2}{\sqrt{11}}}{\frac{2\sqrt{3}}{\sqrt{11}}}\right)=\arctan\left(-\frac{1}{\sqrt{3}}\right).
   $$

5. Since
   $$
   \arctan\left(\frac{1}{\sqrt{3}}\right)=\frac{\pi}{6},
   $$
   it follows that
   $$
   \arctan\left(-\frac{1}{\sqrt{3}}\right)=-\frac{\pi}{6}.
   $$
   However, we need the angle measured in radians between $$0$$ and $$2\pi$$. Adding $$2\pi$$ we obtain
   $$
   -\frac{\pi}{6}+2\pi=\frac{11\pi}{6}.
   $$
   Thus, the angle associated with the **direction of maximum increase** is  
   $$
   \boxed{\frac{11\pi}{6}}.
   $$

6. (For completeness, note that the **direction of maximum decrease** is the direction opposite to the gradient. Its angle is obtained by adding $$\pi$$ to $$\frac{11\pi}{6}$$ and reducing modulo $$2\pi$$:
   $$
   \frac{11\pi}{6}+\pi=\frac{11\pi+6\pi}{6}=\frac{17\pi}{6}.
   $$
   Since $$\frac{17\pi}{6}>2\pi$$, subtract $$2\pi$$ (i.e. $$\frac{12\pi}{6}$$) to obtain
   $$
   \frac{17\pi}{6}-\frac{12\pi}{6}=\frac{5\pi}{6}.
   $$
   Also, the directions for **zero change** are those perpendicular to the gradient. These are given by adding and subtracting $$\frac{\pi}{2}$$ to $$\frac{11\pi}{6}$$. That yields
   $$
   \frac{11\pi}{6}-\frac{\pi}{2}=\frac{11\pi-3\pi}{6}=\frac{8\pi}{6}=\frac{4\pi}{3},
   $$
   and
   $$
   \frac{11\pi}{6}+\frac{\pi}{2}=\frac{11\pi+3\pi}{6}=\frac{14\pi}{6}=\frac{7\pi}{3}.
   $$
   Reducing $$\frac{7\pi}{3}$$ modulo $$2\pi$$,
   $$
   \frac{7\pi}{3}-2\pi=\frac{7\pi-6\pi}{3}=\frac{\pi}{3}.
   $$
   Thus, the zero change directions correspond to $$\theta=\frac{\pi}{3}$$ and $$\theta=\frac{4\pi}{3}$$.)

7. The answer for part (b) is: The **direction of maximum increase** is given by  
   $$
   \boxed{\frac{11\pi}{6}}.
   $$
The angle associated with the direction of maximum increase is $$ \frac{11\pi}{6} $$.

Upstudy AI Solution

Answer

Solution

Beyond the Answer

Related Questions

Latest Calculus Questions